The equation of a circle $C$ is $x^2+y^2-18x-12y+92 = 0$. What is its center $(h, k)$ and its radius $r$ ?
Answer: To find the equation in standard form, complete the square. $(x^2-18x) + (y^2-12y) = -92$ $(x^2-18x+81) + (y^2-12y+36) = -92 + 81 + 36$ $(x-9)^{2} + (y-6)^{2} = 25 = 5^2$ Thus, $(h, k) = (9, 6)$ and $r = 5$.